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Alls you do to find a molar mass is add up all of the atomic masses. Potassium=39.09 Chlorine= 35.453 Oxygen(3)=15.999. So KClO3 would equal 271.65g to a mol?
Then I think to find the number of atoms of each you would take the Atomic Mass * avacodo's number so it would be for example chlorine. 1gCl(35.453/1)(6.022*10^23/1)
However I might be wrong on that part.
What else can I help you with?
How many moles of KClO3 on decomposition give 1 mole of oxygen?
2 grams of Oxygen can be obtained from 5 grams of KClO3 (only if the "CL" means "Cl", which is Chlorine! Remember that only the first letter of the atomic symbol is capitalized.)
Help with stoichiometry show steps How many grams of KClO3 must be decomposed to yield 30 grams of oxygen?
2 KClO3 -> KCL + 3O2 Molar weight of O2 = 32 grams/mole (so close it doesn't matter) 30 grams/32grams/mole = 0.9375 moles Molar weight of KCL = 39+35.5 = 74.5 grams/mole (Want more accuracy? Do it yourself?) now if we have 3 moles of O2 then we have 2 moles of KCl. If we have one mole of O2 then we have 2/3 moles of KCL What ever moles we have of O2 we must multiply it by 2/3 to get the moles of KCl So we have 0.9375moles of O2 x 2/3 = 0.625 moles of KCl So 0.625 moles of KCl x 74.5 grams/mole KCl = 46.5625 grams KCl
How many moles of KCl will be formed from 0.440 moles of KClO3?
2 KClO3 ------ 2KCl + 3O2 so 2 moles of KClO3 produces two mole of KCl. Therefore 0.440 moles of potassium chlorate will produce 0.44 moles of KCl - potassium chloride.
How many milliliters of O2 will form a STP from 55.2 grams of KCLO3?
I assume you mean the decomposition reaction used to produce O2 in lab. 2KCLO3 -> 2KCl + 3O2 find moles O2 55.2 grams KClO3 (1 mole KCLO3/122.55 grams)(3 mole O2/2 mole KClO3) = 0.67564 moles O2 Now, I use the ideal gas law PV = nRT (1 atm)(V) = (0.67564 mol)(0.08206 L*atm/mol*K)(298.15 K) Volume O2 = 16.53 Liters which is 16530 milliliters ( less significant figures )
What volume of oxygen at STP can be profuced by 6.125g of potassium chlorate according to the reaction?
The balanced reaction for the decomposition of potassium chlorate (KClO3) is: 2 KClO3 -> 2 KCl + 3 O2 From the reaction, 2 moles of KClO3 produce 3 moles of O2. Calculate the moles of KClO3 in 6.125g using its molar mass. Convert moles of KClO3 to moles of O2. Use the ideal gas law to find the volume of O2 at STP (22.4 L/mol).
How many moles of KClO3 on decomposition give 1 mole of oxygen?
2 grams of Oxygen can be obtained from 5 grams of KClO3 (only if the "CL" means "Cl", which is Chlorine! Remember that only the first letter of the atomic symbol is capitalized.)
Help with stoichiometry show steps How many grams of KClO3 must be decomposed to yield 30 grams of oxygen?
2 KClO3 -> KCL + 3O2 Molar weight of O2 = 32 grams/mole (so close it doesn't matter) 30 grams/32grams/mole = 0.9375 moles Molar weight of KCL = 39+35.5 = 74.5 grams/mole (Want more accuracy? Do it yourself?) now if we have 3 moles of O2 then we have 2 moles of KCl. If we have one mole of O2 then we have 2/3 moles of KCL What ever moles we have of O2 we must multiply it by 2/3 to get the moles of KCl So we have 0.9375moles of O2 x 2/3 = 0.625 moles of KCl So 0.625 moles of KCl x 74.5 grams/mole KCl = 46.5625 grams KCl
How many moles of KCl will be formed from 0.440 moles of KClO3?
2 KClO3 ------ 2KCl + 3O2 so 2 moles of KClO3 produces two mole of KCl. Therefore 0.440 moles of potassium chlorate will produce 0.44 moles of KCl - potassium chloride.
How many milliliters of O2 will form a STP from 55.2 grams of KCLO3?
I assume you mean the decomposition reaction used to produce O2 in lab. 2KCLO3 -> 2KCl + 3O2 find moles O2 55.2 grams KClO3 (1 mole KCLO3/122.55 grams)(3 mole O2/2 mole KClO3) = 0.67564 moles O2 Now, I use the ideal gas law PV = nRT (1 atm)(V) = (0.67564 mol)(0.08206 L*atm/mol*K)(298.15 K) Volume O2 = 16.53 Liters which is 16530 milliliters ( less significant figures )
How many moles of potassium chlorate must be used to produce 6 moles of oxygen gas?
2KClO3 --> 2KCl + 3O2For every 3 moles of oxygen gas produced, 2 moles of potassium chlorate are used.6 moles O2 * (2 moles KClO3 reacted / 3 moles O2 produced) = 4 moles KClO3
How many moles of oxygen would be produced by 1.2 moles of KClO3?
2 to 3, because of the balanced equation:2 KClO3 --> 2 KCl + 3 O2
What volume of oxygen at STP can be profuced by 6.125g of potassium chlorate according to the reaction?
The balanced reaction for the decomposition of potassium chlorate (KClO3) is: 2 KClO3 -> 2 KCl + 3 O2 From the reaction, 2 moles of KClO3 produce 3 moles of O2. Calculate the moles of KClO3 in 6.125g using its molar mass. Convert moles of KClO3 to moles of O2. Use the ideal gas law to find the volume of O2 at STP (22.4 L/mol).
How many grams of potassium chloride are produced if 25 grams of potassium chlorate decompose into potassium chloride and oxygen?
Oh, dude, chemistry time! So, when 25 grams of potassium chlorate decompose, you get 74.55% potassium chloride and 25.45% oxygen. So, if you do the math, you'd get around 18.64 grams of potassium chloride. But hey, who's counting, right?
How many moles of oxygen are produced from 14 moles of potassium chlorate?
2KClO3 + heat -> 2KCl + 3O2 14 moles KClO3 (3 mole O2/2 mole KClO3) = 21 moles oxygen made This is a common industrial method of producing oxygen.
How many grams of oxygen form when each quantity of reactant completely reacts. 2 KClO3 (s) plus 2 KCl(s) 3O2 (g)?
The balanced chemical reaction equation says that you get 3 moles of O2.m O2 = ( n O2 ) ( M O2 )m O2 = ( 3 mol O2 ) ( 32.00 g O2 / mol O2 = 96 g O2
When 46.0 g KClO3 is heated 12.1 g O2 is formed how many grams of KCl is also formed?
The equation that describes this process is as follows: 2KClO3 ---> 2KCl + 3O2 For every 2 moles of reactants consumed 3 moles of oxygen gas are produced. 3 mol O2 / 2 mol KClO3 = x mol O2 / 12.3 mol KClO3 x = 12.3 mol x 3 mol / 2 mol = 18.45 mol Therefore, 18.5 mol (3 significant figures) of oxygen are produced by the decomposition of 12.3 mol of potassium chlorate
How many moles of oxygen are produced when 7.5 moles of potassium chlorate decompose completely?
For every mole of potassium chlorate that decomposes, three moles of oxygen are produced. Therefore, if 7.5 moles of potassium chlorate decompose, 22.5 moles of oxygen would be produced (7.5 moles x 3).
